A field extension $K / F$ is called finite if $[K : F] < \infty$.
Example 1.4. complex and real numbers, rationals and polynomials [1204]
$\mathbb{C}$ is a field extension of $\mathbb{R}$ and $[\mathbb{C} : \mathbb{R}] = 2$.
$\mathbb{R}$ is a field extension of $\mathbb{Q}$ and $[\mathbb{R} : \mathbb{Q}] = \infty$.
Let $F$ be a field and $p(x)$ be an irreducible polynomial of degree $\geq 1$ in $F[x]$. Then $K = F[x] / (p(x))$ is a field extension of $F$ and $[K : F] = \deg p(x)$.
$F = \mathbb{Q}, p(x) = x^{2} - 2, K = \mathbb{Q}[x] / (x^{2} - 2)$. Then $K \cong \mathbb{Q}[\sqrt{ 2 }] = \{ a + b\sqrt{ 2 } \mid a, b \in \mathbb{Q} \}$.
$[\mathbb{F}_{p}(t), \mathbb{F}_{p}] = \infty$. $1, t, t^{2}, \dots, t^{n}, \dots$ are linearly independent over $\mathbb{F}_{p}$, so $\dim_{\mathbb{F}_{p}}\mathbb{F}_{p}(t) = \infty$.
Let $F \subseteq K \subseteq L$ be fields. Then $[L:F]$ is finite if and only if $[L:K]$ and $[K:F]$ are both finite, and
$$
[L:F] = [L : K][K : F].
$$
$\implies$: Given basis of $L$ over $K$ and $K$ over $F$, denote by $\alpha_{i}, 1 \leq i \leq n$ and $\beta_{j}, 1 \leq j \leq m$ respectively, then we can prove $\alpha_{i}\beta_{j} \in L, \forall i, j$ are linearly independent over $F$. It follows that $[L : F] \geq mn$. So if $[L : F] < \infty$, it must follow that $m < \infty$ and $n < \infty$.
$\impliedby$: Assume now $m, n < \infty$. We can prove that $\alpha_{i}\beta_{j} \forall i, j$ actually spans $L$ over $F$ by showing that every element of $L$ can be written in the form $\sum F \alpha_{i} \beta_{j}$. Then $\alpha_{i} \beta_{j}$ is a basis of $L$ over $F$, hence $[L : F] = mn < \infty$.
When $[L : F] = \infty$, at most one of $[L:K]$ and $[K:F]$ can be finite. For example, $[\mathbb{C} : \mathbb{Q}] = \infty$ and $[\mathbb{R} : \mathbb{Q}] = \infty$ while $[\mathbb{C} : \mathbb{R}] = 2 < \infty$.
Lecture 2. The subfield generated by a subset and simple extensions [l2]
Apr 1, 2026
The subfield generated by a subset and simple extensions
Let $K / F$ be a field extension and $S$ be a subset of $K$. Let
$$
F(S) = \text{the intersection of all subfields of } K \text{ which contains }F\text{ and }S.
$$
One can check that $F(S)$ is a subfield of $K$ and is the smallest subfield of $K$ that contains both $F$ and $S$. We call $F(S)$ the field generated by $S$ over $F$.
When $S = \{ \alpha_{1}, \dots, \alpha_{n} \}$, we write $F(S)$ as $F(\alpha_{1}, \dots, \alpha_{n})$.
Given $S \subseteq K$, taking intersections is a general way to generate a smaller field extension of $F$, which also gives the smallest one that contains both $S$ and $F$.
From this definition, it is hard to compute $F(S)$ because there might be infinite subfields containing $F$ and $S$. So we need a workaround to get $F(S)$.
We know that taking field of fractions gives us a field, if we consider each element of $F(S)$ as a fraction of elements, what do we know about these elements?
They must belong to an integral domain.
They must be composed of elements from $F$ and $S$.
Interestingly, $K$ is a field, so taking fractions of elements from $S$ simply brings another element of $K$, and it also belongs to a subfield of $K$ containing $S$. Actually, to construct a subfield of $K$ containing $S$, we first need to take multiples of any elements of $S$, then take inverses of those multiples, next, take multiples of those inverses and elements of $S$.
Remember that the needed subfields also contain $F$, given a subfield containing $S$, how do we enlarge it so it contains $F$, too? Well, simply multiplying every element by elements of $F$ and taking sums of those multiples is a good way. Recall that $\operatorname{Frac}(F) = F$, hence we have the following proposition:
Proposition 2.2. the field of fractions of $F[S]$ is $F(S)$ [1302]
Given $K / F$ and $S \subseteq K$, let
$$
F[S] = \{ \text{finite sums of }a \cdot \alpha_{1} \alpha_{2} \dots \alpha_{n} \mid a \in F, \alpha_{1}, \dots, \alpha_{n} \in S, n \in \mathbb{Z}_{\geq 0}\}.
$$
Then $F[S]$ is a subring of $K$ and
$$
F(S) = \left\{ \left. \frac{a}{b} \right\vert a, b \in F[S], b \neq 0 \right\}.
$$
($F(S)$ is the fraction field of $F[S]$, and it is exactly the same object as in generated fields.)
Taking multiples ensures $F[S]$ is closed under multiplication. Taking finite sums then ensures $F[S]$ is closed under addition. Moreover, we can take $a = 1$ and $n = 0$, then $F[S]$ contains $1$. It is easy to verify that $F[S]$ is actually an integral domain, thus we can take its field of fractions.
Proposition 2.3. identity on generated fields [1303]
Let $\varphi : F(S) \to F(S)$ be a field homomorphism such that $\varphi(\alpha) = \alpha$ for all $\alpha \in F \cup S$. Then $\varphi = id_{F(S)}$.
By the construction of $F(S)$ from elements of $F \cup S$ in Proposition 1302, obviously $\varphi = id_{F(S)}$.
Definition 2.4. finitely generated, simple extensions and primitive elements [1304]
A field extension $K / F$ is called finitely generated if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
A field extension $K / F$ is called simple if $K = F(\alpha)$ for some $\alpha \in K$. In this case, $\alpha$ is called a primitive element for the field extension $K / F$.
Example 2.5. $F(\theta), \mathbb{F}_p(t) / \mathbb{F}_p, F(x)$ [1305]
Let $p(x)$ be an irreducible element in $F[x]$. Then by Theorem 1206, we know that $K = F[x] / \left( p(x) \right)$ is a field extension of $F$, and $K = F + F\theta + \cdots + F\theta^{n-1}$ with $\theta = \overline{x} \in F[x] / \left( p(x) \right)$. Hence $K = F[\theta] = F(\theta)$, and $K / F$ is a simple extension, and $\theta$ is a primitive element for $K / F$.
$\mathbb{F}_{p}(t) / \mathbb{F}_{p}$ is simple and $t$ is a primitive element for $\mathbb{F}_{p}(t) / \mathbb{F}_{p}$. In this case, $\mathbb{F}_{p}(t) \neq \mathbb{F}_{p}[t]$.
Let $F$ be a field. Then $$F(x) = \left\{ \left. \frac{a(x)}{b(x)} \right\vert a(x), b(x) \in F[x], b(x) \neq 0 \right\}$$ is a field with $$\begin{align*}+ &: \frac{a(x)}{b(x)} + \frac{c(x)}{d(x)} = \frac{a(x) d(x) + b(x) c(x)}{b(x) d(x)} \\ \times &: \frac{a(x)}{b(x)} \cdot \frac{c(x)}{d(x)} = \frac{a(x) c(x)}{b(x) d(x)}\end{align*}.$$$F(x) / F$ is simple and $x$ is a primitive element for $F(x) / F$. $F[x] \neq F(x)$.
(5) $\implies$ (1): Since $F(\alpha) = F[\alpha]$, there exists $c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \in F[\alpha]$ with at least one nonzero coefficient such that it equals $\alpha ^{-1} \in F(\alpha)$. Hence
$$
\begin{align*}
&\alpha ^{-1} = c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \\
\iff &1 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha \\
\iff &0 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha - 1 \in F[\alpha]
\end{align*}
$$
Therefore, $\alpha$ is algebraic over $F$.
(2) $\implies$ (3)(4)(5): Since $\varphi$ is not injective, $\ker \varphi$ is non-trivial. By the First Isomorphism Theorem, $F[x] / \ker\varphi \cong \mathrm{Im} \varphi$. Here, $\mathrm{Im}\varphi$ is a subring of $K$ containing $1$. Now $K$ is a field $\implies$ $\mathrm{Im} \varphi$ is an ID $\implies$ $F[x] / \ker \varphi$ is an ID $\implies$ $\ker\varphi$ is a PID, i.e., $\ker \varphi = \left( p(x) \right)$ for some $p(x) \in F[x]$ $\implies$ $p(x)$ is prime in $F[x]$ $\implies$ $p(x)$ is irreducible in $F[x]$. By Theorem 1206, $F[x] / \ker \varphi$ is a field and
$$
F[x] / \ker\varphi = F + F\overline{x} + \cdots + F\overline{x}^{n-1}.
$$
With $x \mapsto \alpha$ and $p(\alpha) = 0$ for $p(x) \in \ker\varphi$, we obtain
$$
\mathrm{Im}\varphi = F + F\alpha + \cdots + F\alpha^{n-1}.
$$
So $\mathrm{Im}\varphi$ is field extension of $F$ and $[\mathrm{Im}\varphi : F] < \infty$. By definition of $\varphi$, we have $F[\alpha] \subseteq \mathrm{Im}\varphi$. The above implies $\mathrm{Im}\varphi \subseteq F[\alpha]$. So $\mathrm{Im}\varphi = F[\alpha]$. Now $\mathrm{Im}\varphi$ is a field $\implies$ $F[\alpha] = F(\alpha)$. Therefore, $[F(\alpha) : F] < \infty$ and $[F[\alpha] : F] < \infty$.
(4) $\implies$ (1): $[F[\alpha] : F] < \infty$ implies that $F[\alpha] = F + F\alpha + \dots + F \alpha^{n-1}$ for some $n \in \mathbb{Z}_{\ge 0}$. Hence $1 \in F[\alpha]$ can be written as $1 = c_{0} + c_{1}\alpha + \cdots + c_{n-1}\alpha^{n-1}$, i.e., $c_{0} - 1 + c_{1}\alpha + \dots + c_{n-1}\alpha^{n-1} = 0$. So $\alpha$ is algebraic over $F$.
(3) $\implies$ (4): $F(\alpha) \subseteq F[\alpha] \subseteq F$. Then use Theorem 1207.
$K / F$ is finite $\implies$ $K / F$ is algebraic.
Proposition 3.2. minimal polynomial for $\alpha$ over $F$ [1405]
Let $K / F$ be a field extension and $\alpha \in K$ be an element algebraic over $F$. There exists a unique monic irreducible polynomial $m_{\alpha, F}(x) \in F[x]$ such that $m_{\alpha, F}(\alpha) = 0$.
We call the polynomial $m_{\alpha, F}(x)$ the minimal polynomial for $\alpha$ over $F$, and call $\deg m_{\alpha, F}(x)$ the degree of $\alpha$ over $F$.
Existence: let $p(x)$ be the very polynomial in the proof of Proposition 1403. Let $a$ be the leading coefficient of $p(x)$. Then $m_{\alpha, F}(x) = a^{-1} p(x)$. It is obviously monic, and the irreducibility is inherited from $p(x)$.
Uniqueness: use the same notation as in the proof of Proposition 1403. Then $\ker \varphi = \left( p(x) \right) = \left( m_{\alpha, F}(x) \right)$. Suppose there is another monic irreducible polynomial $m_{\alpha, F}'(x) \in F[x]$ such that $m_{\alpha, F}'(\alpha) = 0$. Then $m_{\alpha, F}'(x) \in \left( m_{\alpha, F}(x) \right)$, hence $m_{\alpha, F}'(x) = m_{\alpha, F}(x) q(x)$ for some $q(x) \in F[x]$. Since $m_{\alpha, F}'(x), m_{\alpha, F}(x)$ are monic, $q(x)$ should also be monic. If $\deg q > 0$, then it contradicts irreducibility of $m_{\alpha, F}'(x)$. So $q(x) = 1$, which implies $m_{\alpha, F}' = m_{\alpha, F}$.
Proposition 3.3. $f(\alpha)=0$ if and only if $m_{\alpha, F} \mid f$ [1406]
Let $K / F$ be a field extension and $\alpha \in K$. Suppose that $f(x) \in F[x]$. Then $f(\alpha) = 0$ if and only if $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.
In this case, $f(x)$ is monic and irreducible if and only if $f(x) = m_{\alpha, F}(x)$.
$\implies$: Just use the same argument in the proof of Proposition 1403. We have $f(x) \in \ker \varphi = \left( m_{\alpha, F}(x) \right)$. So $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.
Proposition 3.4. field isomorphisms regarding algebraicity [1407]
Let $K / F$ be a field extension and $\alpha \in K$.
If $\alpha$ is algebraic over $K$, then the map$$F[x] / \left( m_{\alpha, F} \right) \to F(\alpha) (\subseteq K), \qquad \overline{a(x)} \mapsto a(\alpha)$$is a well-defined field isomorphism, and $[F(\alpha) : F] = \deg m_{\alpha, F}(x)$.
If $\alpha$ is transcendental over $F$, then the map$$F(x) \to F(\alpha) (\subseteq K), \qquad \frac{a(x)}{b(x)} \mapsto a(\alpha)b(\alpha)^{-1}$$is a well-defined field isomorphism, and $[F(\alpha) : F] = \infty$.
By the proof of Proposition 1405, the kernel of the following ring homomorphism
$$\varphi : F[x] \to K, \qquad a(x) \mapsto a(\alpha)$$
is $\left( m_{\alpha, F}(x) \right)$. By the First Isomorphism Theorem,
$$F[x] / \left( m_{\alpha, F}(x) \right) \to \mathrm{Im}\varphi, \qquad \overline{a(x)} \mapsto a(\alpha)$$
is a well-defined isomorphism. Moreover, in the proof of Proposition 1403, we have shown $\mathrm{Im}\varphi = F(\alpha)$. And by Theorem 1206, $[F(\alpha) : F] = \deg m_{\alpha, F}(x)$.
Suppose $\frac{a(x)}{b(x)} = \frac{c(x)}{d(x)}$, then
$$a(x)d(x) = b(x)c(x) \implies a(\alpha)d(\alpha) = b(\alpha)c(\alpha) \implies a(\alpha)b(\alpha)^{-1} = c(\alpha)d(\alpha)^{-1}.$$
So the map is well-defined. Also, it is clearly a field homomorphism. By definition of $F(\alpha)$, it is surjective. Denote the map by $\phi$. If $\phi$ is not injective, then there exists $\frac{a(x)}{b(x)}\in F(x)$ such that $a(\alpha) b(\alpha)^{-1} = 0 \implies a(\alpha) = 0$. Then $\alpha$ is algebraic, contradicting $\alpha$ is transcendental. So $\phi$ is an isomorphism.
Now, if $[F(\alpha) : F] < \infty$, then by Proposition 1403, $\alpha$ is algebraic over $F$, a contradiction. So $[F(\alpha) : F] = \infty$.
Theorem 3.5. isomorphism induced in the middle [1408]
Let $K_{1} / F_{1}$ and $K_{2} / F_{2}$ be field extensions, $\varphi : F_{1} \to F_{2}$ be a field isomorphism and let $p_{1}(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0} \in F_{1}[x]$ be an irreducible polynomial, and let $p_{2}(x) = \varphi(a_{n})x^{n} + \cdots + \varphi(a_{1})x + \varphi(a_{0}) \in F_{2}[x]$. If $\alpha \in K_{1}$ is a root of $p_{1}(x)$ and $\beta \in K_{2}$ is a root of $p_{2}(x)$, then
$$
\Psi: F_{1}(\alpha) \to F_{2}(\beta), \quad c_{m}\alpha^{m} + \cdots + c_{1}\alpha + c_{0} \mapsto \varphi(c_{m})\beta^{m} + \cdots + \varphi(c_{1})\beta + \varphi(c_{0})
$$
for $c_{i} \in F_{1}$, is well-defined and is a field isomorphism.
Exegesis 3.5.1. diagram illustrations and a special case [1408A]
General case
$F_{1} = F_{2} = F, K_{1} = K_{2} = K$ implies that if $\alpha, \beta \in K$ are both roots of an irreducible polynomial $p(x) \in F[x]$, then
$$
F(\alpha) \cong F(\beta), \quad a(\alpha) \mapsto a(\beta), \quad a(x) \in F[x]
$$
is well-defined and is an isomorphism.
Special case
Example 3.5.1.1. An example for $F=\mathbb{Q}$ and $K=\mathbb{C}$ [1408AC]
WLOG, assume $p_{1}(x)$ and $p_{2}(x)$ are monic. Then by Proposition 1407,
$$
\begin{align*}
&\varphi_{1} : F_{1}[x] / \left( p_{1}(x) \right) \to F_{1}(\alpha), \qquad a(x) \mapsto a(\alpha), \\
&\varphi_{2} : F_{2}[x] / \left( p_{2}(x) \right) \to F_{2}(\beta), \qquad a(x) \mapsto a(\beta)
\end{align*}
$$
are field isomorphisms. Also, define
$$
\sigma : F_{1}[x] / \left( p_{1}(x) \right) \to F_{2}[x] / \left( p_{2}(x) \right) , \quad c_{m}x^{m} + \cdots + c_{0} \mapsto \varphi(c_{m}) x^{m} + \cdots + \varphi(c_{0}),
$$
which is well-defined and is an isomorphism induced by the isomorphism $\varphi$.
Then $\varphi_{2} \circ \sigma \circ \varphi_{1} : F_{1}(\alpha) \to F_{2}(\beta)$ is a field isomorphism and we have $\Psi = \varphi_{2} \circ \sigma \circ \varphi_{1}^{-1}$.
Proposition 3.6. necessary and sufficient conditions for $K / F$ to be finite or algebraic [14010]
Let $K / F$ be a field extension.
$K / F$ is finite if and only if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ algebraic over $F$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
$K / F$ is algebraic if and only if there exists $S \subseteq K$ such that every $\alpha \in S$ is algebraic over $F$ and $K = F(S)$.
A finite extension requires a finite number of algebraic generators ($\alpha_{1}, \dots, \alpha_{n}$).
An algebraic extension requires a set $S$ of algebraic generators, but $S$ can be infinite.
Not all algebraic extensions are finite. For example, the algebraic closure of $\mathbb{Q}$, denoted as $\overline{\mathbb{Q}}$, is an algebraic extension over $\mathbb{Q}$, but $[\overline{\mathbb{Q}} : \mathbb{Q}] = \infty$.
$\implies$: Suppose $[K : F] = n < \infty$. Let $\alpha_{1}, \dots, \alpha_{n}$ be a basis of $K$ as an $F$-vector space. Then $K = F(\alpha_{1}, \dots, \alpha_{n})$. Moreover, $F(\alpha_{i}) \subseteq K$ implies $[F(\alpha_{i}) : F] \leq [K : F] < \infty$, then by Proposition 1403, $\alpha_{j}$ is algebraic over $F$.
$\impliedby$: By Theorem 1207,
$$[K : F] = [F(\alpha_{1}, \dots, \alpha_{n}) : F] \leq [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] [F(\alpha_{1}, \dots, \alpha_{n-1}) : F].$$
By HW2 Prob 1, $[F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \leq [F(\alpha_{n}) : F]$. So
$$[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{1}, \dots, \alpha_{n-1}) : F]$$
Therefore, by induction we have $[K : F] \leq [F(\alpha_{n}) : F][F(\alpha_{n-1}) : F] \cdots [F(\alpha_{1}) : F]$, in which every $[F(\alpha_{i}) : F] < \infty$.
$\implies$: Take $S = K$.
$\impliedby$: Given $\beta \in K = F(S)$, $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$ for some $\alpha_{i} \in S$. The $\alpha_{i}$’s are algebraic, so $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is finite. Hence $[F(\beta) : F]$ is finite since $F(\beta) \subseteq F(\alpha_{1}, \dots, \alpha_{n})$. Then by Proposition 1403, $\beta$ is algebraic over $F$. Therefore $K / F$ is algebraic.
Let $\alpha, \beta$ be any elements of $E$, it suffices to show that $\alpha + \beta, -\alpha, \alpha \beta, \alpha ^{-1} \in E$. Since they are all in $F(\alpha, \beta)$, it suffices to show that $F(\alpha, \beta) \subseteq E$. By Proposition 1410, $F(\alpha, \beta) / F$ is finite and hence is algebraic. Therefore, $F(\alpha, \beta) \subseteq E$.
Proposition 4.2. $K / F$ and $L / K$ are algebraic $\iff$ $L / F$ is algebraic [14012]
Let $F \subseteq K \subseteq L$ be fields. Then $K / F$ and $L / K$ are algebraic if and only if $L / F$ is algebraic.
$\implies$: Given $\alpha \in L$, it is algebraic over $K$ since $L / K$ is algebraic. So there exist $a_{0}, \dots, a_{n-1} \in K$ such that
$$
a(\alpha) = \alpha^{n} + a_{n-1} \alpha^{n-1} + \cdots + a_{0} = 0.
$$
Hence $\alpha$ is algebraic over $F(a_{0}, \dots, a_{n-1})$, and
$$
[F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})] < \infty.
$$
Since $K / F$ is algebraic, each $a_{i}$ is algebraic over $F$, so $[F(a_{i}) : F] < \infty$ for all $i$. Thus
$$
[F(a_{0}, \dots, a_{n-1}) : F] < \infty.
$$
By Theorem 1207, we write
$$
[F(\alpha, a_{0}, \dots, a_{n-1}) : F] = [F(\alpha, a_{0}, \dots, a_{n-1}) : F(a_{0}, \dots, a_{n-1})][F(a_{0}, \dots, a_{n-1}) : F],
$$
then we can conclude that $[F(\alpha, a_{0}, \dots, a_{n-1}) : F] < \infty$. Since $F \subseteq F(\alpha) \subseteq F(\alpha, a_{0}, \dots, a_{n-1})$, it follows that $[F(\alpha) : F] < \infty$. Then by Proposition 1403, $\alpha$ is algebraic over $F$. Since $\alpha$ is arbitrary, $L / F$ is algebraic.
$\impliedby$: Since $L / F$ is algebraic and $K \subseteq L$, $K / F$ is also algebraic. Moreover, every $\alpha \in L$, there exist $a_{0}, \dots, a_{n-1} \in F$ such that
$$
a(\alpha) = \alpha^{n} + a_{n-1}\alpha^{n-1} + \cdots + a_{0} = 0.
$$
Since $F \subseteq K$, each $a_{i} \in K$, hence $a(x) \in K[x]$, so $\alpha$ is algebraic over $K$. Therefore, $L / K$ is algebraic.
Splitting fields
Given a polynomial $f(x) \in F[x]$ and a field extension $K / F$, we say $f(x)$ splits completely in $K[x]$ or $f(x)$ splits complete over $K$ if $f(x)$ factors completely into a product of linear factors in $K[x]$.
Proposition 4.3. If $f \mid g$, then $g(x)$ splits $\implies$ $f(x)$ splits. [1501]
Let $K / F$ be a field extension and $f(x), g(x) \in F[x]$ such that $f(x) \mid g(x)$. Then $g(x)$ splits completely over $K$ $\implies$ $f(x)$ splits completely over $K$.
Let $F$ be a field and $f(x) \in F[x]$. A field extension $K$ of $F$ is called a splitting field for $f(x)$ (over $F$) if
$f(x)$ splits completely in $K[x]$,
$f(x)$ does not split completely in $E[x]$ for any subfield $E \subsetneq K$ containing $F$. (Or equivalently, if $F \subseteq E \subseteq K$ and $f(x)$ splits completely over $E$, then $E = K$.)
Proposition 4.5. $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n})$ [1503]
Let $K / F$ be a field extension and $f(x) \in F[x]$. Suppose that $f(x)$ splits completely over $K$, i.e., $f(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n})$ with $a, \alpha_{1}, \dots, \alpha_{n} \in K$ (in fact $a \in F$).
$F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x) \in F[x]$ contained in $K$.
If $E \subseteq K$ is also a splitting field of $f(x) \in F[x]$, then $E = F(\alpha_{1}, \dots, \alpha_{n})$.
Clearly, $f(x)$ splits completely in $F(\alpha_{1}, \dots, \alpha_{n})$.
Claim: $F \subseteq E \subseteq K$ and $f(x)$ splits completely over $E$ $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.
Proof of the Claim: Suppose $f(x) = b(x - \beta_{1})\cdots(x - \beta_{n})$ with $b, \beta_{1}, \dots, \beta_{n} \in E$. Then we have
$$
a(x - \alpha_{1}) \cdots (x - \alpha_{n}) = b(x - \beta_{1}) \cdots (x - \beta_{n})
$$
in $K[x]$. Since $K[x]$ is a UFD, after reordering we have $\alpha_{i} = \beta_{i}$ for all $1 \leq i \leq n$. And $F(\beta_{1}, \dots, \beta_{n}) \subseteq E$. Hence $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$.
Suppose $F \subseteq E \subseteq F(\alpha_{1}, \dots, \alpha_{n})$ and $f(x)$ splits completely over $E$, then by our claim, $E = F(\alpha_{1}, \dots, \alpha_{n})$. So $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$.
Since $f(x)$ splits completely over $E$, by our claim, $F(\alpha_{1}, \dots, \alpha_{n}) \subseteq E$. Moreover, $E$ is a splitting field of $f(x)$, then by definition, $E = F(\alpha_{1}, \dots, \alpha_{n})$.
$n = 1$: it is clear that $F[x]$ is a splitting field of $f(x)$.
Assume it holds for $\leq n - 1$. Suppose $f(x) = p(x)a(x)$ for some irreducible $p(x) \in F[x]$ with $\deg p > 0$ and $a(x) \in F[x]$. Then $p(x)$ is prime in $F[x]$, and by Theorem 1206, let $E_{1} = F[x] / \left( p(x) \right)$, then $E_{1}$ is a field extension of $F$ and $\theta = \overline{x} \in E_{1}$ is a root of $p(x)$. Then $\theta$ is also a root of $f(x)$ since $f(x) \in \left( p(x) \right)$. So $f(x) = (x - \theta)f_{1}(x)$ for some $f_{1}(x) \in E_{1}[x]$ and $\deg f_{1} = n - 1$. By induction, there exists a field extension $E / E_{1}$ such that $f_{1}(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n-1})$ with $\alpha_{1}, \dots, \alpha_{n-1} \in E$. Then $f(x) = a(x - \theta)(x - \alpha_{1}) \cdots (x - \alpha_{n-1})$ in $E[x]$, i.e., splits completely in $E[x]$.
By Proposition 1503, $K = F(\theta, \alpha_{1}, \dots, \alpha_{n-1})$ is a splitting field of $f(x)$ in $F[x]$.
By Proposition 1410, $F(\theta, \alpha_{1}, \dots, \alpha_{n-1})$ is a finite extension of $F$.
The splitting field for $x^{2} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt{ 2 })$. $\pm \sqrt{ 2 } \in \mathbb{Q}(\sqrt{ 2 })$. In general, the splitting of $x^{2} + bx + c \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt{ b^{2} - 4c })$.
The splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$.
Exegesis 5.1.2.1. why $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field [1505BA]
Suppose $K$ is a splitting field of $x^{3} - 2$ in $\mathbb{Q}[x]$. The roots of $x^{3} - 2$ are
$$\theta_{1} = \sqrt[3]{ 2 }, \quad \theta_{2} = \sqrt[3]{ 2 } e^{ 2 \pi i / 3 } = \sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2}, \quad \theta_{3} = \sqrt[3]{ 2 } e^{ 4 \pi i / 3 } = -\sqrt[3]{ 2 } \frac{-1 + \sqrt{ -3 }}{2}.$$
They all belong to $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$, hence $K \subseteq \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$. On the other hand, if $K$ contains all the roots, then $\theta_{1} \in K$ and $\theta_{1}^{2}\theta_{2} = -1 + \sqrt{ -3 } \in K \implies \sqrt{ -3 } \in K$, and $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) \subseteq K$. Therefore, $K = \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$.
Exegesis 5.1.2.2. degree of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ [1505BB]
We have
$$
[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}(\sqrt[3]{ 2 })] [\mathbb{Q}(\sqrt[3]{ 2 }) : \mathbb{Q}] = 2 \times 3 = 6.
$$
$[\mathbb{Q}(\sqrt[3]{ 2 }) : \mathbb{Q}] = 3$ is because $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$, which has degree $3$, and then by Proposition 1407.
Similarly, $x^{2} + 3$ is monic and irreducible in $\mathbb{Q}[x]$, so $m_{\sqrt{ -3 }, \mathbb{Q}}(x) = x^{2} + 3$. Moreover, $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{ 2 })$, hence we must have $m_{\sqrt{ -3 }, \mathbb{Q}(\sqrt[3]{ 2 })}(x) \mid m_{\sqrt{ -3 }, \mathbb{Q}}(x)$, which means either $\deg m_{\sqrt{ -3 }, \mathbb{Q}(\sqrt[3]{ 2 })}(x) = 1$ or $2$. The former case implies $x - \sqrt{ -3 } \in \mathbb{Q}\left( \sqrt[3]{ 2 } \right)[x]$, which means $\sqrt{ -3 } \in \mathbb{Q}\left( \sqrt[3]{ 2 } \right)$. But $\sqrt{ -3 }$ is a pure imaginary number, it cannot exist in $\mathbb{Q}\left( \sqrt[3]{ 2 } \right)$. Hence $\deg m_{\sqrt{ -3 }, \mathbb{Q}\left( \sqrt[3]{ 2 } \right)}(x) = 2$, which means $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}(\sqrt[3]{ 2 })] = 2$.
It is easy to see that $\mathbb{Q}(\sqrt{ -3 }), \mathbb{Q}(\theta_{1}), \mathbb{Q}(\theta_{2})$ and $\mathbb{Q}(\theta_{3})$ are all proper subfields of $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ -3 })$ and
$$
[\mathbb{Q}(\theta_{j}) : \mathbb{Q}] = 3 \quad (j = 1, 2, 3), \qquad [\mathbb{Q}(\sqrt{ -3 }) : \mathbb{Q}] = 2.
$$
The splitting field of $x^{n} - 1 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\zeta_{n})$.
The polynomial splits completely in $\mathbb{C}$. Let $\zeta_{n} = e^{ 2\pi i / n } = \cos \left( \frac{2\pi}{n} \right) + i \sin \left( \frac{2\pi}{n} \right) \in \mathbb{C}$. Then $\zeta_{n}$ is a root of $x^{n} - 1$ and $\zeta_{n}^{k}, 0 \leq k \leq n -1$ are distinct roots of $x^{n} - 1$. It follows that
$$
(x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}) \mid x^{n} - 1
$$
in $\mathbb{C}[x]$. By comparing degrees, we know that
$$
x^{n} - 1 = (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}).
$$
Therefore, a splitting field of $x^{n} - 1$ is $\mathbb{Q}(\zeta_{n}, \dots, \zeta_{n}^{n-1}) = \mathbb{Q}(\zeta_{n}) = \mathbb{Q}[\zeta_{n}]$.
If $k$ is coprime to $n$, then $\mathbb{Q}(\zeta_{n}) = \mathbb{Q}(\zeta_{n}^{k})$.
When $n = p$ is prime, we have shown that $\Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = x^{p-1} + \cdots + x + 1$ is irreducible in $\mathbb{Q}[x]$ in MATH 111B. And $\Phi_{p}(\zeta_{p}) = 0$ and $\Phi_{p}(x)$ is monic, so $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x)$. Therefore, $[\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p-1$.
Later, we will show that $[\mathbb{Q}(\zeta_{n}) : \mathbb{Q}] = \lvert (\mathbb{Z} / n\mathbb{Z})^{\times} \rvert$.
All splitting fields of $f(x) \in F[x]$ are isomorphic.
More generally, let $\varphi : F_{1} \xrightarrow{\cong} F_{2}$ be an isomorphism of fields. Given $f_{1}(x) = a_{n}x^{n} + \cdots + a_{1}x + a_{0} \in F_{1}[x]$, let $f_{2}(x) = \varphi(a_{n})x^{n} + \cdots + \varphi(a_{1})x + \varphi(a_{0}) \in F_{2}[x]$. Suppose that $E_{1}$ is a splitting field of $f_{1}(x)$ over $F_{1}$ and $E_{2}$ is a splitting field of $f_{2}(x)$ over $F_{2}$. Then the isomorphism $\varphi$ extends to an isomorphism $\sigma : E_{1} \xrightarrow{\cong} E_{2}$, i.e., there exists $\sigma$ such that the diagram
The induction argument requires the more general statement.
The strategy $F \subseteq F(\alpha) \subseteq K$ with directly handling simple extension $F(\alpha) / F$ and apply induction to $E / F(\alpha)$ will be repeatedly used in several proofs.
Proposition 1503 show that all splitting fields of $f(x)$ contained in a common field extension $K$ of $F$ are the same.
We prove by induction on $\deg f_{1}(x) = n$.
Case $n = 1$: in this case, $E_{1} = F_{1}$ and $E_{2} = F_{2}$. The isomorphism $\sigma$ is simply $\varphi$.
Case $n$: assume the theorem holds for all $\deg f_{1}(x) < n$.
Let $\alpha$ be a root of $f_{1}(x)$ over $E_{1}$, we first find its corresponding root $\beta$ of $f_{2}(x)$ over $E_{2}$ such that $\sigma(\alpha) = \beta$. After that, we establish $F_{1}(\alpha) \cong F_{2}(\beta)$ and write $f_{1}(x) = (x - \alpha)g_{1}(x)$ and $f_{2}(x) = (x - \beta)g_{2}(x)$ with $\deg g_{1}(x) < n$ and $\deg g_{2}(x) < n$, and then use assumption on $g_{1}, g_{2}$ over $F_{1}(\alpha), F_{2}(\beta)$ to get $F_{1}(\alpha)(\alpha_{1}, \dots, \alpha_{n-1}) \cong F_{2}(\beta)(\beta_{1}, \dots, \beta_{n-1})$ where $\alpha_{1}, \dots, \alpha_{n-1}$ are the roots of $g_{1}(x)$ and $\beta_{1}, \dots, \beta_{n-1}$ are the roots of $g_{2}(x)$. And finally claim that $E_{1} = F_{1}(\alpha, \alpha_{1}, \dots, \alpha_{n-1}) = F_{1}(\alpha, \dots, \alpha_{n-1})$ and the same for $E_{2}$.
Proposition 5.3. an upper bound for the degree of the splitting field [1507]
The splitting field of a polynomial of degree $n$ over a base field is of degree at most $n!$.
Let $F$ be the base field and $f(x) \in F[x]$ be the polynomial with $\deg f(x) = n$. Let $\alpha_{1}, \dots, \alpha_{n}$ be the roots of $f(x)$ and $E = F(\alpha_{1}, \dots, \alpha_{n})$ be the splitting field of $f(x)$.
Then $[F(\alpha_{1}) : F]$ has degree at most $n$ since $1, \alpha_{1}, \dots, \alpha_{1}^{n}$ is not linearly independent ($f(\alpha_{1}) = 0$). Moreover, by Homework 2, $[F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})] < [F(\alpha_{1}) : F] \leq n \implies [F(\alpha_{1}, \alpha_{2}): F(\alpha_{1})] \leq n-1$. If we keep doing so, we will finally arrive at
$$
\begin{align*}
[E : F] &= [F(\alpha_{1}, \dots, \alpha_{n}) : F(\alpha_{1}, \dots, \alpha_{n-1})] \cdots [F(\alpha_{1}, \alpha_{2}) : F(\alpha_{1})][F(\alpha_{1}) : F] \\
&\leq 1 \cdots \cdot (n-1)n \\
&= n!
\end{align*}
$$
Algebraic closure
Proposition 5.4. equivalent definitions of algebraically closed fields [1601]
For a field $\Omega$, the following statements are equivalent:
Every non-constant polynomial in $\Omega[x]$ splits completely in $\Omega$.
Every non-constant polynomial in $\Omega[x]$ has a root in $\Omega$.
The irreducible polynomials in $\Omega[x]$ are those of degree $1$.
If $K / \Omega$ is an algebraic extension, then $K = \Omega$.
If $K / \Omega$ is a finite extension, then $K = \Omega$.
3 $\implies$ 4: Take $\alpha \in K$ and consider $m_{\alpha, \Omega}(x)$. It must be $m_{\alpha, \Omega}(x) = x - \alpha \in \Omega[x]$ since it is irreducible and monic, hence $\alpha \in \Omega$. So we conclude that $K = \Omega$.
5 $\implies$ 1: Take any $f(x) \in \Omega[x]$ and let $K$ be its splitting field. Then by Proposition 1507, $[K : \Omega]$ is at most $n!$ where $n = \deg f(x)$, hence it is finite. So $K = \Omega$. Then we can conclude that every non-constant polynomial in $\Omega[x]$ splits completely in $\Omega$.
Proposition 6.1. extensions of field embedding [1603]
Let $K / F$ be an algebraic field extension and $\Omega$ be an algebraically closed field. Every field embedding $\varphi : F \to \Omega$ extends to $K$, i.e. there exists a field embedding $\psi : K \to \Omega$ such that the diagram
commutes.
We prove by induction on $[K / F] = n$. The initial case $n = 1$ implies $K = F$ and hence $\varphi = \psi$. Assume the theorem holds for $n-1$. Pick any $\alpha \in K - F$, we consider $m_{\alpha, F}(x) \in F[x]$. Our strategy is to use Theorem 1506 to deduce it from $n$ to $n-1$ and then use the assumption.
Clearly $\varphi(m_{\alpha, F}(x)) \in \Omega[x]$. Since $\Omega$ is algebraically closed, $\varphi(m_{\alpha, F}(x))$ has a root $\beta \in \Omega$. Moreover, $\varphi(m_{\alpha, F(x)})$ is monic and irreducible because $m_{\alpha, F}(x)$ is monic and irreducible. So $m_{\beta, \Omega}(x) = \varphi(m_{\alpha, F}(x))$. Now we can apply Theorem 1506 and obtain an isomorphism $\varphi' : F(\alpha) \to \Omega$such that $\varphi'|_{F} = \varphi$. Then, by our induction hypothesis, we can extend $\varphi'$ to $\psi : K \to \Omega$ since
$$
[K : F(\alpha)] = \frac{[K : F]}{[F(\alpha) : F]}
$$
and we have $[F(\alpha) : F] > 1$ (because $\alpha \not\in F$).
Case 2: $K / F$ is infinite.
In this case, we use Zorn’s Lemma and the proved case that $K / F$ is finite for assistance.
Let $S$ be the set of pairs $(L, \phi)$ where $L$ is a subfield of $K$ containing $F$ with $\phi : L \to \Omega$ a field embedding with $\phi|_{F} = \varphi$. First, $S$ is not empty since $(F, \operatorname{id}_{F}) \in S$. Second, we define an order on $S$ by
$$
(L, \phi) \leq (L', \phi') \iff L \subset L' \text{ and } \phi'|_{L} = \phi.
$$
Suppose $C \subseteq S$ is a chain. We show that $C$ has an upper bound in $S$. Let
$$
E = \bigcup_{(L, \phi) \in C}L.
$$
$E$ is indeed a subfield of $K$. Given $\alpha \in E$, there exists $(L, \phi) \in C$ such that $\alpha \in L$. Set $\sigma(\alpha) = \phi(\alpha)$. We show that $\sigma(\alpha)$ does not depend on the choice of $(L, \phi)$. If $(L', \phi')$ is another element in $C$ satisfying $\alpha \in L'$, then $(L', \phi') \leq (L, \phi)$ or $(L, \phi) \leq (L', \phi')$ because $C$ is a chain. If it is the former case, then $L' \subseteq L$ and $\phi |_{L'} = \phi'$, which implies $\phi'(\alpha) = \phi(\alpha)$. The latter case is almost the same. Therefore, for every $\alpha \in E$, we have defined a unique $\sigma(\alpha) \in \Omega$. So $\sigma : E \to \Omega, \alpha \mapsto \phi(\alpha)$ is a well-defined map. Moreover, $\sigma |_{F} = \varphi$ and $(E, \sigma) \in S$ and is an upper bound of $C$. By Zorn’s Lemma, $S$ has a maximal element $(L, \phi)$.
We then show that $L = K$ by contradiction. If $L \subsetneq K$, pick any $\alpha \in K - L$, then $L(\alpha) / L$ is finite. By the case of finite extension proved above, $\phi : L \to \Omega$ extends to $\phi': L(\alpha) \to \Omega$. Then $(L, \phi) \leq (L(\alpha), \varphi')$ and $(L, \phi) \neq (L(\alpha), \phi')$. This contradicts the maximality of $(L, \phi)$. Therefore, $L = K$ and $\phi$ is the desired extension of $\varphi$.
Let $F$ be a field. A field $\Omega$ containing $F$ is called an algebraic closure of $F$ if it is algebraic over $F$ and every (non-constant) polynomial in $F[x]$ splits completely over $\Omega$.
$\implies$: Take any non-constant $f(x) \in \Omega[x]$. By Theorem 1504, $f(x)$ has a splitting field $\Omega'$ and $\Omega' / \Omega$ is finite thus algebraic by Proposition 1404. Then by Proposition 14012, since $\Omega' / \Omega$ and $\Omega / F$ are algebraic, $\Omega' / F$ is algebraic too. Hence for any $\alpha \in \Omega'$, there exists $a(x) \in F[x]$ such that $a(\alpha) = 0$. Now, since $\Omega$ is an algebraic closure of $F$, $a(x)$ splits completely in $\Omega$, which implies $\alpha \in \Omega$. Therefore, $\Omega' = \Omega$. So every non-constant $f(x) \in \Omega[x]$ splits completely over $\Omega$, $\Omega$ is algebraically closed.
Clearly $F' / F$ is algebraic. Take any $f(x) \in F[x]$, $f(x)$ splits completely over $\Omega$. By Proposition 1503, $F(\alpha_{1}, \dots, \alpha_{n})$ is a splitting field of $f(x)$, where $f(x) = a(x - \alpha_{1})\cdots(x - \alpha_{n}), a \in F, \alpha_{i} \in \Omega$. Hence each $\alpha_{i}$ is algebraic over $F$ and belongs to $F'$, which implies that $f(x)$ splits completely over $F'$. Therefore, $F'$ is an algebraic closure of $F$.
$\alpha \in \Omega$ is algebraic over $F$ implies it is algebraic over $K$, so $\alpha \in K'$, which means $F' \subseteq K'$. By Proposition 14012, $K' / F$ is algebraic. So every $\beta \in K'$ is algebraic over $F$, hence $\beta \in F'$, it follows that $K' \subseteq F'$. So $K' = F'$.
Proposition 6.5. uniqueness of algebraic closures up to isomorphisms [1607]
Let $F$ be a field.
$F$ has an algebraic closure.
If $L, L'$ are both algebraic closure of $F$, then there exists a field isomorphism $\varphi : L \to L'$ such that $\varphi|_{F} = \operatorname{id}_{F}$.
Apply Proposition 1603 to $L / F$ and the natural field embedding $F \to L'$, then there exists a field embedding $\varphi : L \to L'$ such that $\varphi |_{F} = \operatorname{id}_{F}$. By Proposition 1605, $L$ and $L'$ are both algebraically closed. Then $L$ is algebraically closed $\implies$ $\varphi(L)$ is algebraically closed since $L \cong \varphi(L)$ (recall that a field homomorphism is either $0$ or its kernel is $0$). By Proposition 14012, $L' / F$ is algebraic $\implies$ $L' / \varphi(L)$ is algebraic. By part 4 of Proposition 1601, $L' = \varphi(L)$ since $L'$ is algebraically closed.
We often denote an algebraic closure by $\overline{F}$.
If $K / F$ is an algebraic extension, then $\overline{K}$ is an algebraic closure of $F$.
In particular, $\overline{F} \cong \overline{K}$. The converse is not true, i.e. $\overline{K} \cong \overline{F}$ does not imply $K / F$ is algebraic, because it can happen $K / F$ is not algebraic and $K \cong F$, for example, $F = \mathbb{Q}(x_{1}, x_{2}, \dots, x_{m}, \dots), K = \mathbb{Q}(x_{0}, x_{1}, \dots, x_{n}, \dots)$.
Lecture 7. Separable extensions and field embeddings [l7]
Apr 20, 2026
I did’t go to both Lecture 6 and Lecture 7, so I don’t know where it stopped at the end of Lecture 6. But whatever, let’s just pretend that it splits here.
Let $F$ be a field and $\overline{F}$ be an algebraic closure of $F$. Let $f(x) \in F[x]$. A root $\alpha \in \overline{F}$ of $f(x)$ is said to be of multiplicity $m$ if
$$
(x - \alpha)^{m} \mid f(x), \text{ and } (x - \alpha)^{m+1} \nmid f(x) \text{ in } \overline{F}[x].
$$
A root of multiplicity $1$ is called a simple root. A root of multiplicity $\geq 2$ is called a multiple root.
Proposition 7.2. roots of an irreducible have the same multiplicity [1702]
Let $f \in F[x]$ be an irreducible polynomial. Then all roots of $f(x)$ in $\overline{F}$ have the same multiplicity.
Let $\alpha, \beta \in \overline{F}$ be roots of $f(x)$ with multiplicities $a, b$ respectively. Suppose $a < b$. Since $f(x)$ is monic and irreducible in $F[x]$, $m_{\alpha, F}(x) = m_{\beta, F}(x) = f(x)$. Therefore, by Theorem 1408, there exists a field isomorphism
$$
\varphi : F(\alpha) \to F(\beta)
$$
such that $\varphi(\alpha) = \beta$ and $\varphi(r) = r$ for all $r \in F$.
Over $F(\alpha)$, $f(x) = m_{\alpha, F}(x) = (x - \alpha)^{a}g(x)$ for some $g(x) \in F(\alpha)[x]$ such that $(x - \alpha) \nmid g(x)$. Over $F(\beta)$, $f(x) = m_{\beta, F}(x) = (x - \beta)^{b}h(x)$ for some $h(x) \in F(\beta)[x]$ such that $(x - \beta) \nmid h(x)$. Then
$$
\begin{align*}
\varphi \left( f(x) \right) &= (x - \varphi(\alpha))^{a} \varphi \left( g(x) \right) \\
&= (x- \beta)^{a} \varphi \left( g(x) \right) \\
&= (x - \beta)^{b}h(x).
\end{align*}
$$
It follows that $\varphi \left( g(x) \right) = (x - \beta)^{b-a}h(x)$. Since $b > a$, $(x - \beta) \mid \varphi \left( g(x) \right)$. Apply $\varphi ^{-1}$ on $\varphi \left( g(x) \right)$, it turns out that
$$
g(x) = \varphi ^{-1} \varphi \left( g(x) \right) = (x - \alpha)^{b-a} \varphi ^{-1} \left( h(x) \right).
$$
Hence $(x - \alpha) \mid g(x)$ in $F(\alpha)$, contradicting the fact that $(x - \alpha) \nmid g(x)$. So we must have $a = b$, i.e. the multiplicities of $\alpha$ and $\beta$ are equal. Because $\alpha, \beta$ are arbitrary, all roots of $f(x)$ in $\overline{F}$ have the same multiplicity.
Let $F$ be a field and $\overline{F}$ be an algebraic closure of $F$.
A polynomial in $F[x]$ is called separable if all its roots in $\overline{F}$ are simple. A polynomial which is not separable is called inseparable.
Let $K / F$ be a field extension and $\alpha \in K$ be an element algebraic over $F$. We say $\alpha$ is separable over $F$ if its minimal polynomial over $F$ is separable.
An algebraic extension $K / F$ is called separable if every element in $K$ is separable over $F$. An algebraic extension $K / F$ is called inseparable if it is not separable.
Let $f(x) \in F[x]$. Its derivative $f'(x)$ is defined as
$$
f'(x) = na_{n}x^{n-1} + \cdots + 2a_{2}x + a_{1}.
$$
One can check that for $f(x), g(x) \in F[x]$, we have
$$
\left( f(x) + g(x) \right)' = f'(x) + g'(x), \qquad \left( f(x)g(x) \right)' = f'(x)g(x) + f(x)g'(x).
$$
$\alpha \in \overline{F}$ is a multiple root of $f(x) \in F[x]$ if and only if it is a root of both $f(x)$ and $f'(x)$.
$f(x) \in F[x]$ is separable if and only if $\gcd \left( f(x), f'(x) \right) = 1$.
An irreducible polynomial in $F[x]$ is separable if and only if $f'(x) \neq 0$.
If $F$ is a field of characteristic $0$, then every irreducible polynomial in $F[x]$ is separable and every field extension of a field of characteristic $0$ is separable.
Claim: $\alpha \in \overline{F}$ is a common root of $f(x)$ and $f'(x)$ if and only if $\alpha$ is a root of $d(x)$. Then, just follow part 1.
Let’s prove the claim.
$\impliedby$: It is clear that $d(\alpha) = 0 \implies f(\alpha) = f'(\alpha) = 0$.
$\implies$: Since $F[x]$ is a PID, there exists $a(x), b(x) \in F[x]$ such that $a(x)f(x) + b(x)f'(x) = d(x)$. Then $d(\alpha) = a(\alpha)f(\alpha) + b(\alpha)f'(\alpha) = 0$.
A direct proposition of part 2.
By part 3.
$\operatorname{ch}(F) = 0 \implies f'(x) \neq 0$, then use part 3. For every $K / F$ and $\alpha \in K$, by part 4, $m_{\alpha, F}(x)$ is separable, so $\alpha$ is separable over $F$, hence $K / F$ is separable.
Lecture 8. Separable extensions and field embeddings (cont.) [l8]
If $F$ is a field of characteristic $p$ and $a \in F$, then $x^{p} - a$ is inseparable. To prove that, we divide it into two cases:
If $a$ is a $p$-th power in $F$, let $\alpha \in \overline{F}$ such that $\alpha^{p} = a$, then $\alpha \in F$.Now $x^{p} - a = x^{p} - \alpha^{p} = (x - \alpha)^{p}$, $\alpha$ has multiplicity $p > 1$. So just by definition, $x^{p} - a$ is inseparable.
If $a$ is NOT a $p$-th power in $F$, then $\alpha \not\in F$. By the HW problem in section 1.4, $m_{\alpha, F}(x) = x^{p} - a$. Now taking the derivative of $x^{p} - a$ gives $px^{p-1} = 0$ and then by part 3 of Proposition 1704, $x^{p} - a$ is inseparable.
It follows that $F(\alpha) / F$ is inseparable in both cases.
Consider $F = \mathbb{F}_{p}(t) = \left\{ \frac{a(t)}{b(t)} \mid a(t), b(t) \in \mathbb{F}_{p}[t], b(t) \neq 0 \right\}$. $t \in F$ is not a $p$-th power in $F$ since if $t = \left( \frac{a(t)}{b(t)} \right)^{p}$ for some $a(t), b(t) \in \mathbb{F}_{p}[t]$ with $b(t) \neq 0$, then we would have
$$
t \cdot \left( b(t) \right)^{p} = \left( a(t) \right)^{p}.
$$
Suppose $\deg a(t) = m, \deg b(t) = n$. Then comparing degrees of both sides gives $np + 1 = mp \implies 1 = p(m - n)$. Since $p, m, n$ are integers and $p > 1$ (bc $p=1 \implies t \in \mathbb{F}_{p}$), this is impossible. Therefore, $t$ cannot be a $p$-th power in $F$.
So again we have $x^{p} - t$ is irreducible in $F[x]$. Moreover, by Proposition 1407, $F(\alpha) \cong F[x] / \left( m_{\alpha, F}(x) \right) = F[x] / \left( x^{p} - t \right)$ if $\alpha \in \overline{F}$ is a root of $x^{p} - t$. This implies $F[x] / (x^{p} - t)$ is an inseparable extension of $F$ by our previous discussion.
Let $K, L$ be field extensions of $F$. An $F$-embedding $\varphi : K \to L$ is a field embedding such that $\varphi |_{F} = \operatorname{id}_{F}$.
Proposition 8.3. the number of $F$-embeddings [1707]
Let $K / F$ be a finite field extension. Then
$$
\# \{ F\text{-embeddings } \varphi : K \to \overline{F} \} \leq [K : F],
$$
and the equality holds if and only if $K / F$ is separable.
Trivial case: $[K : F] = 1$. In this case, $K = F$, the proposition is obviously true.
Assume the statement holds for all field extensions of degree $\leq n - 1$. Given a field extension $K / F$ with $[K : F] = n$, take $\alpha \in K - F$, by Proposition 17010,
$$
\begin{align*}
&\# \{ F\text{- embeddings } \varphi : K \to \overline{F} \} \\
= &\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \cdot \#\{ F(\alpha)\text{-embeddings }\varphi : K \to \overline{F(\alpha)} = \overline{F} \}.
\end{align*}
$$
By Proposition 1709,
$$
\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \} \leq [F(\alpha) : F]
$$
$[K : F(\alpha)] < n$, so by the induction hypothesis,
$$
\# \{ F(\alpha)\text{-embeddings } \varphi : K \to \overline{F} \} \leq [K : F(\alpha)].
$$
Therefore, by Theorem 1207,
$$
\#\{ F\text{-embeddings } \varphi : K \to \overline{F} \} \leq [F(\alpha) : F][K : F(\alpha)] = [K : F].
$$
The equality holds if and only if the equality holds for
$$
\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \} \leq [F(\alpha) : F].
$$
So only if
$$
\begin{align*}
&\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} = [F(\alpha) : F] = \deg m_{\alpha, F}(x) \\
\iff &\text{all roots of }m_{\alpha, F}(x)\text{ are simple} \\
\iff &\alpha\text{ is separable over }F.
\end{align*}
$$
Since $\alpha \in K - F$ is arbitrarily chosen, it implies that the equality holds if and only if all $\alpha \in K - F$ is separable over $F$, so if and only if $K / F$ is separable.
Remark 8.4. another definition of separable extension [1708]
In some literature, a finite extension $K / F$ is defined to be separable if
$$
\# \{ F\text{-embeddings } \varphi : K \to \overline{F} \} = [K : F].
$$
To prove Proposition 1707, we need the following two propositions:
$\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} \leq [F(\alpha) : F]$ and the equality holds if and only if $\alpha$ is separable over $F$.
The map
$$\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \to \overline{F}, \qquad \varphi \mapsto \varphi(\alpha)$$
is injective since the image of $\varphi$ is completely determined by $\varphi(\alpha)$. To associate it with the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, suppose $\varphi(\alpha) = \beta \in \overline{F}$. Then since $\varphi |_{F} = \operatorname{id}_{F}$ and $m_{\alpha, F}(x) \in F[x]$, we have $\varphi(m_{\alpha, F}(\alpha)) = m_{\alpha, F}(\beta) = \varphi(0) = 0$, hence $\beta$ is a root of $m_{\alpha, F}(x)$ in $\overline{F}$. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \leq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.
Conversely, given a root $\gamma$ of $m_{\alpha, F}(x)$ in $\overline{F}$, then the kernel of the field homomorphism $F[x] \to \overline{F}, a(x) \mapsto a(\gamma)$ is $\left( m_{\alpha, F}(x) \right)$. Thus we obtain a field embedding
$$\tau : F[x] / \left( m_{\alpha, F}(x) \right) \to \overline{F}, \qquad \overline{a(x)} \mapsto a(\gamma).$$
Moreover, by Proposition 1407,
$$\sigma : F[x] / \left( m_{\alpha, F}(x) \right) \to F(\alpha), \qquad \overline{a(x)} \mapsto a(\alpha)$$
is a field isomorphism. Then we can check that the field homomorphism $\tau \circ \sigma ^{-1} |_{F} = \operatorname{id}_{F}$ and $\tau \circ \sigma ^{-1} (\alpha) = \gamma$, so $\tau \circ \sigma ^{-1}$ is the desired field embedding. Thus $\# \{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \geq \# \{ \text{roots of }m_{\alpha, F}(x) \text{ in }\overline{F} \}$.
By $\# \{ \text{roots of }m_{\alpha, F}(x) \text{ in } \overline{F} \} \leq \deg m_{\alpha, F}(x) = [F(\alpha) : F]$, and the equality holds if and only if $\alpha$ is separable over $F$.
$\implies$: Suppose $L / F$ is separable. Given $\alpha \in K$, we have that $\alpha \in L$ is hence separable over $F$, so $K / F$ is separable. To show that $L / K$ is separable, consider $m_{\alpha, F}(x)$. Since $\alpha$ is separable over $F$, all roots of $m_{\alpha, F}(x)$ are simple. Moreover, $m_{\alpha, K}(x) \mid m(\alpha, F)(x)$ in $K[x]$, so all roots of $m_{\alpha, K}(x)$ are simple as well $\implies$ $\alpha$ is separable over $K$. Hence $L$ is separable over $K$.
$\impliedby$: Suppose that $K / F$ and $L / K$ are both separable.
Case 1: $[L : F] < \infty$. By Proposition 1707,
$$
\#\{ F\text{-embeddings }\varphi : K \to \overline{F} \} = [K : F], \, \#\{ K\text{-embeddings } \varphi : L \to \overline{K} \} = [L : K].
$$
Then Proposition 17010 implies
$$
\#\{ F\text{-embeddings }\varphi : L \to \overline{F} \} = [K : F][L : K] = [L : F].
$$
By Proposition 1707, this implies that $L / F$ is separable.
Proposition 9.1. $K / F$ is separable iff $K = F(S)$ and $S$ is separable over $F$ [17012]
Let $K$ be an algebraic extension of $F$. Then $K / F$ is separable if and only if there exists $S \subseteq K$ such that $K = F(S)$ and every element in $S$ is separable over $F$.
Case 1: $[K : F] < \infty$. Use induction on $[K : F]$. The inertial case $[K : F] = 1$ is obviously true. Assume the statement holds for all field extensions of degree $\leq n - 1$. Suppose $[K : F] = n, K = F(S)$ and every element in $S$ is separable over $F$. Take any $\alpha \in S \setminus F$, $\alpha$ is separable over $F$ $\implies$ $\#\{ F\text{-embeddings }\varphi : F(\alpha) \to \overline{F} \} = [F(\alpha) : F]$ by Proposition 17010. Moreover, by induction hypothesis, $[K : F(\alpha)] < [K : F] = n$ (bc $\alpha \not\in F$), it follows that $K / F(\alpha)$ is separable. So by Proposition 1707, $\#\{ F(\alpha)\text{-embeddings} \varphi : K \to F(\alpha) \} = [K : F(\alpha)]$. Thus by Proposition 17010,
$$
\begin{align*}
&\#\{ F\text{-embeddings } \varphi : K \to \overline{F} \} \\
= &\#\{ F\text{-embeddings } \varphi : F(\alpha) \to \overline{F} \} \cdot \#\{ F(\alpha)\text{-embeddings } \varphi : K \to \overline{F(\alpha)} = \overline{F} \} \\
= &[F(\alpha) : F][K : F(\alpha)] \\
= &[K : F].
\end{align*}
$$
So again, by Proposition 1707, $K / F$ is separable.
Case 2: $[K : F] = \infty$. Since $K = F(S)$, given any $\beta \in K$, there exists $\alpha_{1}, \dots, \alpha_{n} \in S$ such that $\beta \in F(\alpha_{1}, \dots, \alpha_{n})$. Now Case 1 $\implies$ $F(\alpha_{1}, \dots, \alpha_{n}) / F$ is separable, hence $\beta$ is separable over $F$. Therefore, $K / F$ is separable.
Finite fields
Theorem 9.2. the splitting field of $x^{p^n}-x \in \mathbb{F}_p[x]$ is $\mathbb{F}_{p^n}$ [1801]
Let $p$ be a prime number and $n$ be a positive integer.
Let $\mathbb{F}_{p^{n}} \subseteq \overline{\mathbb{F}}_{p}$ be the splitting field of $x^{p^{n}} - x \in \mathbb{F}_{p}[x]$. Then $\lvert \mathbb{F}_{p^{n}} \rvert = p^{n}$ and $\mathbb{F}_{p^{n}}$ is the unique subfield of $\overline{\mathbb{F}}_{p}$ pf cardinality $p^{n}$.
If $F$ is a field with $\lvert F \rvert = p^{n}$, $F \cong \mathbb{F}_{p^{n}}$.
Let $S \subseteq \overline{\mathbb{F}}_{p}$ be the set of roots of $x^{p^{n}} - x$. We first show that $S$ is a subfield of $\overline{\mathbb{F}}_{p}$, then we show that $\lvert S \rvert = p^{n} = \deg (x^{p^{n}} - x)$, this suffices to show that $x^{p^{n}} - x$ is separable.
First, $0 \in S$. Given $\alpha, \beta \in S$, we have $(\alpha - \beta)^{p^{n}} - (\alpha - \beta) = \alpha^{p^{n}} - \beta^{p^{n}} - (\alpha - \beta) = 0$. Hence $\alpha - \beta \in S \implies (S, +)$ is a subgroup of $(\overline{\mathbb{F}}_{p}, +)$. Next, $(\alpha\beta)^{p^{n}} - \alpha\beta = \alpha^{p^{n}}\beta^{p^{n}} - \alpha\beta = 0 \implies \alpha\beta \in S$ and $(\alpha ^{-1})^{p^{n}} - \alpha ^{-1} = (\alpha^{p^{n}})^{-1} - \alpha ^{-1} = 0 \implies \alpha ^{-1} \in S$ shows that $S$ is a subfield of $\overline{\mathbb{F}}_{p}$. Moreover, $\mathbb{F}_{p^{n}} = \mathbb{F}_{p}[S] = S$. This is because any $\alpha \in \mathbb{F}_{p}$ satisfies $\alpha^{p^{n}} - \alpha = \alpha^{p^{n} - 1}\alpha - \alpha = 1 \cdot \alpha - \alpha = 0$ by Fermat’s little theorem. Hence $\alpha \in S$ $\implies$ $\mathbb{F}_{p}[S] = S$.
Now since $(x^{p^{n}} - x)' = p^{n}x^{p^{n} - 1} - 1 = -1$, we have $\gcd(x^{p^{n}} - x, (x^{p^{n}} - x)') = 1$. By Proposition 1704, $x^{p^{n}} - x$ is separable over $\mathbb{F}_{p}$, hence $\lvert S \rvert = p^{n}$.
Then we prove the uniqueness of $\mathbb{F}_{p^{n}}$. Suppose that $F$ is a subfield of $\overline{\mathbb{F}}_{p}$ with $\lvert F \rvert = p^{n}$. Then the multiplicative group $F^{\times}$ is abelian of order $p^{n} - 1$. By Lagrange’s Theorem, we know that $\alpha^{p^{n} - 1} = 1$ for all $\alpha \in F^{\times}$. Hence every element in $F$ is a root of $x^{p^{n}} - x$, hence $F \subseteq \mathbb{F}_{p^{n}}$. Since they have the same cardinality, $F = \mathbb{F}_{p^{n}}$.
Let $F_{0}$ be $F$’s prime field. By the definition of $[F : F_{0}]$, which is $\dim_{F_{0}} F$, we have $\lvert F \rvert = \lvert F_{0} \rvert^{[F:F_{0}]}$. Hence $\lvert F_{0} \rvert$ is a prime that divides $p^{n}$ $\implies$ $\lvert F_{0} \rvert = p \implies F_{0} \cong \mathbb{F}_{p}$. Let $\varphi : F_{0} \to \overline{\mathbb{F}}_{p}$ be defined by $F_{0} \xrightarrow{\cong} \mathbb{F}_{p} \hookrightarrow \overline{\mathbb{F}}_{p}$. By [Proposition 1603], $\varphi$ extends to a field embedding $\psi : F \to \overline{\mathbb{F}}_{p}$. Then $\mathrm{Im}\psi \cong F$ is a subfield of $\overline{\mathbb{F}}_{p}$ and $\lvert \mathrm{Im}\psi \rvert = \lvert F \rvert = p^{n}$. By (1), we know that $\mathrm{Im}\psi = \mathbb{F}_{p^{n}}$. Therefore, $F \cong \mathbb{F}_{p^{n}}$.
Homework: Show that $\mathbb{F}_{2}[x] / (x^{3} + x + 1) \cong \mathbb{F}_{2}[y] / (y^{3} + y^{2} + 1)$ and find an explicit isomorphism.
Let $F$ be a field of characteristic $0$. Given finite field extension $K / F$ and $\alpha \in K$, let $m_{\alpha, F}(x) \in F[x]$ be the minimal polynomial of $\alpha$ over $F$. Then $m_{\alpha, F}(x)$ is irreducible. By (3) of Proposition 1704, $m_{\alpha, F}(x)$ is separable. Hence $\alpha$ is separable over $F$. So $K / F$ is separable $\implies$ $F$ is perfect.
Let $F$ be a finite field with $\operatorname{ch}(F) = p$ and $\lvert F \rvert = p^{n}$ for some integer $n$. By the proof of Theorem 1801 ,$\alpha = \alpha^{p^{n}} = (\alpha^{p^{n-1}})^{p}$ for every $\alpha \in F$. In particular, every element in $F$ is a $p$-th power in $F$. By (3), $F$ is perfect.
Suppose that $F = F^{p}$. Let $K / F$ be a finite extension and $\alpha \in K$. Let $m_{\alpha, F}(x)$ be the minimal polynomial of $\alpha$ over $F$. We show that $m_{\alpha, F}(x)$ is separable by contradiction. Suppose the opposite. Then by (2) of Proposition 1704, $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) \neq 1$, then $\gcd \left( m_{\alpha, F}(x), m_{\alpha, F}'(x) \right) = m_{\alpha, F}(x)$ because $m_{\alpha, F}(x)$ is irreducible. It follows that $m_{\alpha, F}(x) \mid m'_{\alpha, F}(x)$. We deduce that $m_{\alpha, F}'(x) = 0$. Otherwise, $\deg m_{\alpha, F}'(x) < \deg m_{\alpha, F}(x)$, so $m_{\alpha, F}(x)$ cannot divide $m_{\alpha, F}'(x)$. It follows from the definition of derivative of polynomials that$$m_{\alpha, F}'(x) = 0 \implies m_{\alpha, F}(x) = a_{m}x^{p^{m}} + a_{m-1}x^{p^{m-1}} + \cdots + a_{1}x^{p} + a_{0}.$$The condition $F = F^{p}$ implies that there exists $b_{0}, \dots, b_{m} \in F$ such that $a_{j} = b_{j}^{p}$. Then$$\begin{align*} m_{\alpha, F}(x) &= b_{m}^{p}x^{p^{m}} + b_{m-1}^{p}x^{p^{m-1}} + \cdots + b_{1}^{p}x^{p} + b_{0}^{p} \\ &= \left( b_{m}x^{p^{m-1}} + b_{m-1}x^{p^{m-1}} + \cdots + b_{1}x + b_{0} \right)^{p},\end{align*}$$and $m_{\alpha, F}(x)$ is not irreducible. We get a contradiction. Hence, $m_{\alpha, F}(x)$ is separable, and $\alpha$ is separable over $F$, so $K / F$ is separable and $F$ is perfect.
Galois theory
Lecture 10. Groups of automorphisms of fields [l10]
An automorphism of $F$ is an isomorphism from $F$ to itself. We denote by $\mathrm{Aut}(F)$ the set of all automorphisms of $F$.
An automorphism $\sigma$ of $F$ is said to fix $\alpha \in F$ if $\sigma(\alpha) = \alpha$, and is said to fix a subset $S \subseteq F$ if $\sigma(\alpha) = \alpha$ for all $\alpha \in S$.
For every field $F$, $\operatorname{id}_{F}$ is an automorphism of $F$.
If $\sigma$ is an automorphism of $F$, then $\sigma(1) = 1$, and $\sigma(1 + 1 + \cdots + 1) = 1 + 1 + \cdots + 1$. Hence, $\sigma$ fixes the prime field of $F$.
Let $F$ be a field of characteristic $p$ and $\mathrm{Fr} : F \to F$ be the Frobenius endomorphism (sending $a \in F$ to $a^{p}$). $\mathrm{Fr}$ is a field homomorphism, so it is injective. It is surjective if and only if every element in $F$ is a $p$-th power, i.e. if and only if $F$ is perfect by Proposition 1902. Thus, $\mathrm{Fr}$ is an automorphism if and only if the field is perfect.
Given $H \leq \mathrm{Aut}(K / F)$, define$$K^{H} = \{ \alpha \in K \mid \sigma(\alpha) = \alpha \text{ for all } \sigma \in H \}.$$Then $K^{H}$ is a subfield of $K$ and $H \leq \mathrm{Aut}(K / K^{H})$. $K^{H}$ is called the fixed field of $H$.
Let $K / F$ be an algebraic field extension. If $\alpha \in K$ with $m_{\alpha, F}(x)$ and $\sigma \in \mathrm{Aut}(K / F)$, then $\varphi(\alpha)$ is a root of $m_{\alpha, F}(x)$. Therefore, we have the map
$$
\mathrm{Aut}\left( F(\alpha) / F \right) \to \{ \text{root of }m_{\alpha, F}(x) \text{ in }F(\alpha) \}, \qquad \sigma \mapsto \sigma(\alpha).
$$
The map is injective because an automorphism $\sigma \in \mathrm{Aut}\left( F(\alpha) / F \right)$ is completely determined by $\sigma(\alpha)$. Next, we show that it is also surjective. If $\beta \in F(\alpha)$ is a root of $m_{\alpha, F}(x)$, then $F(\beta) = F(\alpha)$ ($[F(\alpha) : F(\beta)] = \frac{[F(\alpha) : F]}{F(\beta) : F} = \frac{\deg m_{\alpha, F}(x)}{\deg m_{\beta, F}(x)} = 1$), and
$$
\sigma: F(\alpha) \to F(\beta) = F(\alpha), \qquad a(\alpha) \mapsto a(\beta) \, \forall a(x) \in F[x]
$$
defines an automorphism of $F(\alpha)$ fixing $F$ by Theorem 1408. Thus, $\beta = \sigma(\alpha)$ belongs to the image of the map. This shows that it is surjective.
Example 2106 (1) can be viewed as a special case of this. The minimal polynomial of $\sqrt{ 2 }$ over $\mathbb{Q}$ is $x^{2} - 2$, and the set $\mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }) / \mathbb{Q} \right)$ is in bijection with the set of roots of $x^{2} - 2$ in $\mathbb{Q}(\sqrt{ 2 })$ which is $\{ \pm \sqrt{ 2 } \}$.
$F = \mathbb{Q}$ and $K = \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$, which is the splitting field of $x^{3} - 2$ over $\mathbb{Q}$ with $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = 6$ (part 2 of Example 1505). We show that $\mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}$.
Denote the roots of $x^{3} - 2$ by $\theta_{1}, \theta_{2}, \theta_{3}$. Define the map as
$$
\begin{equation}
\mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \to S_{3}, \qquad \sigma \mapsto \tau_{\sigma}
\end{equation}
$$
such that for $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right)$, the permutation $\tau_{\sigma} \in S_{3}$ is defined as $\tau_{\sigma}(i) = j$ if and only if $\sigma(\theta_{i}) = \theta_{j}$.
First we need to check that the map $(1)$ is indeed a group homomorphism. Then simply noticing that $\theta_{1}, \theta_{2}, \theta_{3}$ generate $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ since $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field of $x^{3} - 2$ can show that $(1)$ is injective. To show the surjectivity, it remains to prove that every $\tau \in S_{3}$ corresponds to a valid $\sigma \in \mathrm{Aut}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right)$ such that $\tau = \sigma_{\tau}$.
Let $F$ be a field and $A$ be a subset of $F[x]$. An algebraic field extension $K$ of $F$ is called a splitting field for $A$ (over $F$) if
every $f(x) \in A$ splits completely in $K[x]$
if $F \subseteq L \subseteq K$ and every $f(x) \in A$ splits completely in $L[x]$, then $L = K$.
Remark 11.3. special case that $A$ is finite [2203]
If $A$ is a finite subset of $F[x]$, say $A = \{ f_{1}(x), \dots, f_{n}(x) \}$. Then a field extension $K$ of $F$ is a splitting field for $A$ over $F$ if and only if $K$ is a splitting field for
$$
f(x) = f_{1}(x) f_{2}(x) \cdots f_{n}(x) \in F[x].
$$
Proposition 11.4. existence and uniqueness of the splitting field for $A$ [2204]
Let $F$ be a field and $A$ be a subset of $F[x]$.
A splitting field for $A$ over $F$ exists.
If $K$ and $L$ are both splitting fields for $A$ over $F$, then there exists field isomorphism $\varphi : K \to L$ with $\varphi|_{F} = \operatorname{id}_{F}$.
One can check that $F(S)$ is a splitting field for $A$ over $F$, where $S \subseteq \overline{F}$ is the subset consisting of roots of the polynomials in $A$. (cf. Proposition 1503)
It suffices to show that if $K$ is a splitting field for $A$ over $F$, then there exists a field isomorphism $\varphi : K \to F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$. Apply Proposition 1603 tothen we obtain a field embedding $\varphi : K \to \overline{F}$. Every polynomial splits completely over $K$ $\implies$ $S \subseteq \varphi(K)$ $\implies$ $F(S) \subseteq \varphi(K)$ $\implies$ $\varphi ^{-1}(F(S)) \subseteq K$. By (2) of Definition 2202, since $K$ is a splitting field for $A$ over $F$, it follows that $\varphi ^{-1}(F(S)) = K$ $\implies$ $F(S) = \varphi(K)$. Moreover, $\varphi$ is injective and $\varphi |_{F} = \operatorname{id}_{F}$. So $\varphi$ is a field isomorphism that maps $K$ to $F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$.
Proposition 11.5. equivalent definitions of normal extensions [2205]
Let $K / F$ be an algebraic extension. TFAE:
$K / F$ is normal.
Any irreducible polynomial in $F[x]$ that has a root in $K$ splits completely in $K$.
There exists a subset $S \subseteq K$ such that $K = F(S)$ and $m_{\alpha, F}(x)$ splits completely in $K$ for every $\alpha \in S$.
$K$ is the splitting field for a subset of $F[x]$.
Fix an algebraic closure of $F$ with $F \subseteq K \subseteq \overline{F}$. (For example, $\overline{F} = \overline{K}$.) Any $F$-embedding $\varphi : K \to \overline{F}$ satisfies $\varphi(K) = K$.
Fix an algebraic closure of $F$ with $F \subseteq K \subseteq \overline{F}$. Any $F$-embedding $\varphi : K \to \overline{F}$ satisfies $\varphi(K) \subseteq K$.
Remark 11.6. the textbook definition for normal extensions [2206]
The textbook uses (5) of Proposition 2205 as the definition of normal extensions.
Lecture 12. Normal extensions and Galois extensions (cont.) [l12]
Fix an algebraic closure of $F$ with $F \subseteq K \subseteq \overline{F}$. (By (2) of Proposition 1606, one can simply take $\overline{F} = \overline{K}$). Then we have the following bijective map$$\begin{align*}\mathrm{Aut}(K / F) &\to \left\{ F\text{-embeddings }\varphi : K \to \overline{F} \text{ s.t. } \varphi(K) = K\right\}, \\ \sigma &\mapsto \varphi_{\sigma} \text{ defined as } \varphi_{\sigma}(\alpha) = \sigma(\alpha).\end{align*}$$
If $K / F$ is finite, then$$\lvert \mathrm{Aut}(K / F) \rvert \leq \# \{ F\text{-embeddings from } K \text{ to } \overline{F} \}$$and the equality holds if and only if $K / F$ is normal.
Example 12.2. $\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}$ [2208]
For all $a \in \mathbb{F}_{p}^{\times}$, $x^{p} - x + a$ is irreducible in $\mathbb{F}_{p}[x]$.
Let $\alpha \in \overline{\mathbb{F}_{p}}$ be a root of $x^{p} - x + a$. To show $x^{p} - x + a$ is irreducible, it suffices to show that $m_{\alpha, \mathbb{F}_{p}}(x) = x^{p} - x + a$. Moreover, since $m_{\alpha, \mathbb{F}_{p}}(x) \mid (x^{p} - x + a)$, it suffices to show that $\deg m_{\alpha, \mathbb{F}_{p}}(x) \geq p$. We also know that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] = \deg m_{\alpha, \mathbb{F}_{p}}(x)$, so we only need to prove that $[\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}] \geq p$. This can be done by showing that
$$
p \leq \lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \leq \#\{ \mathbb{F}_{p}\text{-embeddings }\varphi : \mathbb{F}_{p}(\alpha) \to \overline{\mathbb{F}_{p}} \} \leq [\mathbb{F}_{p}(\alpha) : \mathbb{F}_{p}].
$$
$\operatorname{Fr}$ is a good instance of $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$. If $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}, \operatorname{Fr}, \dots, \operatorname{Fr}^{p-1}$ are different, then of course $\lvert \mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p}) \rvert \geq p$ for that $\mathrm{Aut}(\mathbb{F}_{p}(\alpha) / \mathbb{F}_{p})$ is a group. Let us check it on $\alpha$. First, since $\alpha$ is a root of $x^{p} - x + a$, we have
$$
\operatorname{Fr}(\alpha) = \alpha^{p} = \alpha - a.
$$
Then
$$
\operatorname{Fr}^{2}(\alpha) = \operatorname{Fr}(\alpha - a) = \operatorname{Fr}(\alpha) - \operatorname{Fr}(a) = \alpha - a - a^{p} = \alpha - 2a.
$$
Using induction on $n$, we can prove that $\operatorname{Fr}^{n}(\alpha) = \alpha - na$. Therefore, $\operatorname{id}_{\mathbb{F}_{p}(\alpha)}(\alpha), \operatorname{Fr}(\alpha), \dots, \operatorname{Fr}^{p-1}(\alpha)$ are differentiable, so are those automorphisms.
Example 12.3. $\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ [2209]
$\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ is not normal. $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$ does not split completely in $\mathbb{Q}(\sqrt[3]{ 2 })$ since the other two roots $\sqrt[3]{ 2 } \frac{-1 \pm \sqrt{ -3 }}{2}$ do not belong to $\mathbb{Q}(\sqrt[3]{ 2 })$.
$\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is normal.
Use (3) of Proposition 2205: $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$ and $m_{\sqrt{ -3 }, \mathbb{Q}}(x) = x^{2} + 3$ both split completely over $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$.
Use (4) of Proposition 2205: $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ is the splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$.
If $K / F$ is a Galois extension, we call the group $\mathrm{Aut}(K / F)$ the Galois group of $K / F$ and denote it by $\mathrm{Gal}(K / F)$.
Example 12.6. $\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ [22012]
$\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ is not Galois because it is not normal as seen in (1) of Example 2209.
In (2) of Exmaple 2209, we know that $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is normal. It is also separable because $\operatorname{ch}(\mathbb{Q}) = 0$ and by Proposition 1704. Thus $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is a Galois extension. Moreover, by (4) of Example 2106, we have$$\mathrm{Gal}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}.$$
Proposition 12.7. necessary and sufficient conditions for finite $K / F$ to be Galois [22013]
A finite extension $K / F$ is Galois if and only if $\lvert \mathrm{Aut}(K / F) \rvert = [K : F]$.
Fix $\overline{F} = \overline{K}$. Then $F \subseteq K \subseteq \overline{F}$.
$\implies$: Let $A = \{ m_{\alpha, F}(x) \in F[x] \mid \alpha \in K \}$. First, we show that $K$ is the splitting field of $A \subseteq F[x]$. Let $S = \{ \text{roots of } \overline{F} \text{ of polynomials in } A \}$. Then $F(S)$ is the splitting field for $A$. We need to show $K = F(S)$. Given $\alpha \in K$, by the definition of $A$, $m_{\alpha, F}(x) \in A$ and by the definition of $S$, $\alpha$ is a root of $m_{\alpha, F}(x) \in A$, so $\alpha \in S$. Thus, $K \subseteq S \subseteq F(S)$. Conversely, if $\beta \in S$, then $\beta$ is a root of $m_{\alpha, F}(x)$ for some $\alpha \in K$. Since $K / F$ is normal, $K$ contains all the roots of $m_{\alpha, F}(x)$ in $\overline{F}$, so $K$ contains $\beta$. This shows that $S \subseteq K$, so $F(S) \subseteq K$. We have proved $K = F(S)$, the splitting field for $A \subseteq F[x]$.
Also, because $K / F$ is separable, we know that every $\alpha \in K$ is separable over $F$ and $m_{\alpha, F}(x)$ is separable. Hence $A$ is a subset of separable polynomials in $F[x]$.
$\impliedby$: Suppose $A$ is the subset of separable polynomials in $F[x]$ and $K$ is the splitting field for $A \subseteq F[x]$. We show that $K / F$ is both normal and separable. Proposition 2205 $\implies$ $K / F$ is normal. By the proof of Proposition 2204, $K \cong F(S)$ where $S \subseteq \overline{F}$ is the set of roots of polynomials in $A$. The polynomials in $A$ are separable $\implies$ every element in $S$ is separable over $F$. Then Proposition 17012 $\implies$ $K = F(S)$ is a separable extension of $F$.
Suppose $f_{n} \to f$ uniformly on a set $E$ in a metric space. Let $x$ be a limit point of $E$, and suppose that
$$
\lim_{ t \to x } f_{n}(t) = A_{n} \quad (n = 1, 2, 3, \dots).
$$
Then $\{ A_{n} \}$ converges, and
$$
\lim_{ t \to x } f(t) = \lim_{ n \to \infty } A_{n}.
$$
In other words, the conclusion is that
$$
\lim_{ t \to x } \lim_{ n \to \infty } f_{n}(t) = \lim_{ n \to \infty } \lim_{ t \to x } f_{n}(t).
$$
A family $\mathscr{F}$ of complex functions $f$ defined on a set $E$ in a metric space $X$ is said to be equicontinuous on $E$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that
$$
\lvert f(x) - f(y) \rvert < \epsilon
$$
whenever $d(x, y) < \delta, x \in E, y \in E$, and $f \in \mathscr{F}$. Here $d$ denotes the metric of $X$.
It is clear that every member of an equicontinuous family is uniformly continuous.
Theorem 1.4. pointwise boundedness of $\mathscr{C}$ in a countable set [br723]
If $\{ f_{n} \}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_{n} \}$ has a subsequence $\{ f_{n_{k}} \}$ such that $\{ f_{n_{k}}(x) \}$ converges for every $x \in E$.
Let $E = \{ x_{i} \}$. Since $\{ f_{n}(x_{1}) \}$ is a bounded sequence on $\mathbb{C}$, we can find a subsequence that converges to a point of $\mathbb{C}$, denote it by $S_{1} = \{f_{1, k}(x_{1})\}$.
Now, given $\{ f_{1, k}(x_{2}) \}$, a bounded sequence on $\mathbb{C}$, we can again construct a subsequence that converges, denote it by $S_{2} = \{ f_{2, k}(x_{2}) \}$.
Similarly, we do the same for every $x_{i}$, and thus obtain sequences $S_{1}, S_{2}, S_{3}, \dots$, which we represent by the array
$$
\begin{align*}
&S_{1}: \ f_{1,1} \; f_{1, 2} \; f_{1, 3} \; f_{1, 4} \; \dots \\
&S_{2}: \ f_{2,1} \; f_{2, 2} \; f_{2, 3} \; f_{2, 4} \; \dots \\
&S_{3}: \ f_{3,1} \; f_{3, 2} \; f_{3, 3} \; f_{3, 4} \; \dots \\
&\dots \dots \dots \dots \dots \dots \dots \dots
\end{align*}
$$
Then, $\{ f_{n,n} \}$ is the desired subsequence of $\{ f_{n} \}$.
Theorem 1.5. compactness and uniform convergence implies equicontinuity [br724]
If $K$ is a compact metric space, if $f_{n} \in \mathscr{C}(K)$ for $n = 1, 2, 3, \dots$, and if $\{ f_{n} \}$ converges uniformly on $K$, then $\{ f_{n} \}$ is equicontinuous on $K$.
$\{ f_{n} \}$ converges uniformly on $K$ $\implies$ for all $\epsilon > 0$, there exists an integer $N > 0$ such that for any $n, m > N$ we have $\lVert f_{n} - f_{m} \rVert < \epsilon$.
Pick any $k > N$. $f_{k} \in \mathscr{C}(K) \implies$ $f_{k}$ continuous on $K$. Now $K$ is compact $\implies$ $f_{k}$ is uniformly continuous on $K$. So there exists $\delta > 0$ such that for any $x, y \in K$ with $d(x, y) < \delta$, we have $\lvert f_{k}(x) - f_{k}(y) \rvert < \epsilon$. Then for any $n > N$,
$$
\lvert f_{n}(x) - f_{n}(y) \rvert = \lvert (f_{n}(x) - f_{k}(x)) + (f_{k}(x) - f_{k}(y)) + (f_{k}(y) - f_{n}(y)) \rvert < 3\epsilon.
$$
For $1 \leq n \leq N$, $f_{n}$ is uniformly continuous, just take $\delta' = \min_{1 \leq i \leq N}\delta_{i} > 0$, and our final choice of $\delta^*$ is $\min \{ \delta, \delta' \}$, which is positive.
$\{ f_{n} \}$ is equicontinuous on $K$ $\implies$ for all $\epsilon > 0$, there exists $\delta > 0$ such that any $x, y \in K$ with $d(x, y) < \delta$ implies $\lvert f_{n}(x) - f_{n}(y) \rvert < \epsilon$ for all $n$.
For all $x \in K$, let $B_{x}$ denote the neighborhood of $x$ with radius $\delta$. Then $\{ B_{x} \}$ is an open cover of $K$. Since $K$ is compact, there exists a finite set of points $\{ p_{1}, \dots, p_{m} \}$ such that $\{ B_{p_{i}} \}$ is a finite subcover of $K$.
Now consider $B_{p_{i}}$. Let $M_{i}$ be an upper bound of $\{ \lvert f_{n}(p_{i}) \rvert \}$. Then for all $x \in B_{p_{i}}$, $|f_{n}(x)| < |f_{n}(p_{i})| + \epsilon \leq M_{i} + \epsilon$ for all $n$. Hence $M_{i} + \epsilon$ is a global upper bound of $\{ f_{n} \}$ on $B_{p_{i}}$.
Take $M = \max_{i} M_{i} + \epsilon < \infty$, then for all $x \in K$, $\lvert f_{n}(x) \rvert < M$ for all $n$. So $\{ f_{n} \}$ is uniformly bounded on $K$.
Choose any countable dense set $E$ on $K$. Then there exists a subsequence $\{ f_{n_{k}} \}$ of $\{ f_{n} \}$ such that $\{ f_{n_{k}} \}$ converges at every point of $E$. Let $\{ g_{k} \}$ denote $\{ f_{n_{k}} \}$.
Use the same $\{ B_{p_{i}} \}$ as in the first part. Pick any $x \in E$, then $x \in B_{p_{i}}$ for some $i$. Then $\{ g_{k}(x) \}$ converges, i.e., there exists a positive integer $N_{i}$ such that any $n, m > N_{i}$ means $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon / 3$. Now for any $y \in B_{p_{i}} \cap E$, we have that for any $n, m > N_{i}$,
$$\lvert g_{n}(y) - g_{m}(y) \rvert = \lvert \left( g_{n}(y) - g_{n}(x) \right) + \left( g_{n}(x) - g_{m}(x) \right) + \left( g_{m}(x) - g_{m}(y) \right) \rvert < \epsilon$$
Now choose $N = \max_{i} N_{i} < \infty$. For any $x \in E$, $n, m > N$ implies $\lvert g_{n}(x) - g_{m}(x) \rvert < \epsilon$. Hence $\{ g_{k} \}$ is uniformly convergent. Therefore, $\{ f_{n} \}$ contains a uniformly convergent subsequence.
A mapping $A$ of a vector space $X$ into a vector space $Y$ is said to be a linear transformation if
$$
A (\mathbf{x}_{1} + \mathbf{x}_{2}) = A \mathbf{x}_{1} + A \mathbf{x}_{2}, \quad A(c \mathbf{x}) = cA \mathbf{x}
$$
for all $\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x} \in X$ and all scalars $c$.
We often writes $A \mathbf{x}$ insteand of $A(\mathbf{x})$ if $A$ is linear.
Linear transformations of $X$ into $X$ are often called linear operators on $X$. If $A$ is a linear operator on $X$ which is one-to-one and onto, we say that $A$ is invertible.
Definition 2.2. $L(X, Y)$, products and norms [br96]
Let $L(X, Y)$ be the set of all linear transformations of the vector space $X$ into the vector space $Y$. Instead of $L(X, X)$, we shall simply write $L(X)$.
If $X, Y, Z$ are vector spaces, and if $A \in L(X, Y)$ and $B \in L(Y,Z)$, we define their product $BA$ to be the composition of $A$ and $B$:
$$(BA)\mathbf{x} = B(A\mathbf{x}) \qquad (\mathbf{x} \in X).$$
Then $BA \in L(X, Z)$.
For $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, define the norm $\lVert A \rVert$ of $A$ to be the sup of all numbers $\lvert A\mathbf{x} \rvert$, where $\mathbf{x}$ ranges over all vectors in $\mathbb{R}^{n}$ with $\lvert \mathbf{x} \rvert \leq 1$.
Observe that the inequality
$$\lvert A\mathbf{x} \rvert \leq \lVert A \rVert \lvert \mathbf{x} \rvert $$
holds for all $\mathbf{x} \in \mathbb{R}^{n}$. Also, if $\lambda$ is such that $\lvert A\mathbf{x} \rvert \leq \lambda \lvert \mathbf{x} \rvert$ for all $\mathbf{x} \in \mathbb{R}^{n}$, then $\lVert A \rVert \leq\lambda$.
If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, then $\lVert A \rVert < \infty$ and $A$ is a uniformly continuous mapping of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$.
If $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $c$ is a scalar, then$$\lVert A + B \rVert \leq \lVert A \rVert + \lVert B \rVert , \qquad \lVert cA \rVert = \lvert c \rvert \lVert A \rVert.$$With the distance between $A$ and $B$ defined as $\lVert A - B \rVert$, $L(\mathbb{R}^{n}, \mathbb{R}^{m})$ is a metric space.
If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $B \in L(\mathbb{R}^{m}, \mathbb{R}^{n})$, then$$\lVert BA \rVert \leq \lVert B \rVert \lVert A \rVert.$$
Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ be the standard basis in $\mathbb{R}^{n}$ and suppose $\mathbf{x} = \sum c_{i}\mathbf{e}_{i}$. Now $\lvert \mathbf{x} \rvert \leq 1 \implies \lvert c_{i} \rvert \leq 1$ for all $i$. Then
$$\lvert A \mathbf{x} \rvert = \left\lvert \sum c_{i} A \mathbf{e}_{i} \right\rvert \leq \sum \lvert c_{i} \rvert \lvert A \mathbf{e}_{i} \rvert \leq \sum \lvert A \mathbf{e}_{i} \rvert$$
so that
$$\lVert A \rVert \leq \sum_{i=1}^{n} \lvert A \mathbf{e}_{i} \rvert < \infty. $$
Each $A \mathbf{e}_{i}$ is a specific vector in $\mathbb{R}^{m}$, and hence $\lvert A \mathbf{e}_{i} \rvert < \infty$.
Since $\lvert A \mathbf{x} - A \mathbf{y} \rvert \leq \lVert A \rVert \lvert \mathbf{x} - \mathbf{y} \rvert$ if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, we see that $A$ is uniformly continuous.
Omitted.
Omitted.
Theorem 2.4. the set of invertible linear operators [br98]
Let $\Omega$ be the set of all invertible linear operators on $\mathbb{R}^{n}$.
If $A \in \Omega, B \in L(\mathbb{R}^{n})$, and$$\lVert B - A \rVert \cdot \lVert A^{-1} \rVert < 1,$$then $B \in \Omega$.
$\Omega$ is an open subset of $L(\mathbb{R}^{n})$, and the mapping $A \to A^{-1}$ is continuous on $\Omega$.
Suppose $\{ \mathbf{x}_{1}, \dots, \mathbf{x}_{n} \}$ and $\{ \mathbf{y}_{1}, \dots, \mathbf{y}_{m} \}$ are bases of vector spaces $X$ and $Y$, respectively. Then every $A \in L(X, Y)$ determines a set of numbers $a_{ij}$ such that
$$
A \mathbf{x}_{j} = \sum_{i=1}^{m} a_{ij} \mathbf{y}_{i} \qquad (1 \leq j \leq n).
$$
We represent $A$ by an $m$ by $n$ matrix:
$$
[A] = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m 1} & a_{m 2} & \cdots & a_{mn}
\end{bmatrix}
$$
$[A]$ depends not only on $A$ but also on the choice of bases in $X$ and $Y$.
If $\mathbf{x} = \sum_{j} \mathbf{x}_{j}$, then the Schwarz inequality shows that
$$
\lvert A \mathbf{x} \rvert^{2} = \sum_{i} \left( \sum_{j} a_{ij} c_{j} \right)^{2} \leq \sum_{i} \left( \sum_{j} a_{ij}^{2} \cdot \sum_{j} c_{j}^{2} \right) = \sum_{i, j} a_{ij}^{2} \lvert \mathbf{x} \rvert ^{2}.
$$
Thus
$$
\lVert A \rVert \leq \sqrt{ \sum_{i, j} a_{ij}^{2} }.
$$
Moreover, if we replace $A$ by $B - A$, where $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, and view each $a_{ij}$ as continuous functions of a single parameter, then we have the following:
If $S$ is a metric space, if $a_{11}, \dots, a_{mn}$ are real continuous functions on $S$, and if, for each $p \in S$, $A_{p}$ is the linear transformation of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ whose matrix has entries $a_{ij}(p)$, then the mapping $p \mapsto A_{p}$ is a continuous mapping of $S$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.
The theory on continuity of single-variable scalar valued functions can be directly generalized to multi-variable vector-valued functions except left and right limits.
If $f: (a, b) \to \mathbb{R}^{1}$ and if $f'(x_{0})$ exists for $x_{0} \in (a, b)$, i.e., the limit
$$
\lim_{ x \to x_{0} } \frac{f(x) - f(x_{0})}{x - x_{0}}
$$
exists, then let $h = x - x_{0}$, we have the following
$$
\lim_{ h \to 0 } \frac{f(x_{0} + h) - f(x_{0}) - hf'(x_{0})}{h} = 0.
$$
For a scalar-valued single variable function, the existence of the derivative is equivalent to say that
$$
\frac{f(x_{0} + h) - \left( f(x_{0}) + hf'(x_{0}) \right)}{h} \to 0
$$
with respect to $h$.
To generalize to muti-variable vector-valued functions:
Replace single variable to multi-variable: $(x, y) = (x_{0}, y_{0}) + \vec{h}$. We have replaced $h$ by $\lVert \vec{h} \rVert$.$$\vec{v} = (v_{1}, v_{2}), \qquad \lVert \vec{v} \rVert = \sqrt{ v_{1}^{2} + v_{2}^{2} }$$
Then the needed limit is$$\frac{f(x, y) - \left[ f(x_{0}, y_{0}) + \frac{\partial f}{\partial x}(x_{0}, y_{0})(x - x_{0}) + \frac{\partial f}{\partial y}(x_{0}, y_{0})(y - y_{0}) \right]}{\lVert \vec{h} \rVert }$$where$$f(x, y) - \left[ f(x_{0}, y_{0}) + \frac{\partial f}{\partial x}(x_{0}, y_{0})(x - x_{0}) + \frac{\partial f}{\partial y}(x_{0}, y_{0})(y - y_{0}) \right]$$is the error between linear part and $f(x, y)$.
Ensure the existence of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.
Suppose $E$ is an open set in $\mathbb{R}^{n}$, $\mathbf{f}$ maps $E$ into $\mathbb{R}^{m}$, and $\mathbf{x} \in E$. If there exists a linear transformation $A$ of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ such that
$$
\lim_{ \mathbf{h} \to 0 } \frac{\lvert \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A \mathbf{h} \rvert }{\lvert \mathbf{h} \rvert } = 0,
$$
then we say $\mathbf{f}$ is differentiable at $\mathbf{x}$, and we write
$$
\mathbf{f}'(\mathbf{x}) = A.
$$
If $\mathbf{f}$ is differentiable at every $\mathbf{x} \in E$, we say that $\mathbf{f}$ is differentiable in $E$.
Theorem 3.3. uniqueness of $\mathbf{f}'(\mathbf{x})$ [br912]
Suppose $E$ and $\mathbf{f}$ are as in Definition 911, $\mathbf{x} \in E$, and
$$
\lim_{ \mathbf{h} \to 0 } \frac{\lvert \mathbf{f}(\mathbf{x} + \mathbf{h}) - \mathbf{f}(\mathbf{x}) - A \mathbf{h} \rvert }{\lvert \mathbf{h} \rvert } = 0
$$
holds with $A = A_{1}$ and with $A = A_{2}$. Then $A_{1} = A_{2}$.
Suppose $E$ is an open set in $\mathbb{R}^{n}$, $\mathbf{f}$ maps $E$ into $\mathbb{R}^{m}$, $\mathbf{f}$ is differentiable at $\mathbf{x}_{0} \in E$, $\mathbf{g}$ maps an open set containing $\mathbf{f}(E)$ into $\mathbb{R}^{k}$, and $\mathbf{g}$ is differentiable at $\mathbf{f}(\mathbf{x}_{0})$. Then the mapping $\mathbf{F}$ of $E$ into $\mathbb{R}^{k}$ defined by
$$
\mathbf{F}(\mathbf{x}) = \mathbf{g}\left( \mathbf{f}(\mathbf{x}) \right)
$$
is differentiable at $\mathbf{x}_{0}$, and
$$
\mathbf{F}'(\mathbf{x}_{0}) = \mathbf{g}'\left( \mathbf{f}(\mathbf{x}_{0}) \right) \mathbf{f}'(\mathbf{x}_{0}).
$$
Consider $\mathbf{f}: E \to \mathbb{R}^{m}$ where $E \subset \mathbb{R}^{n}$, Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ and $\{ \mathbf{u}_{1}, \dots, \mathbf{u}_{m} \}$ be the standard bases of $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$. The components of $\mathbf{f}$ are the real functions $f_{1}, \dots, f_{m}$ defined by
$$
\mathbf{f}(\mathbf{x}) = \sum_{i=1}^{m} f_{i}(\mathbf{x}) \mathbf{u}_{i} \qquad (\mathbf{x} \in E),
$$
or equivalently, by $f_{i}(\mathbf{x}) = \mathbf{f}(\mathbf{x}) \cdot \mathbf{u}_{i}, 1 \leq i \leq m$.
For, $\mathbf{x} \in E$, $1 \leq i \leq m, 1 \leq j \leq n$, we define
$$
(D_{j}f_{i})(\mathbf{x}) = \lim_{ t \to 0 } \frac{f_{i}(\mathbf{x} + t \mathbf{e}_{j}) - f_{i}(\mathbf{x}) }{t},
$$
provided the limit exists. Writing $f_{i}(x_{1}, \dots, x_{n})$ in place of $f_{i}(\mathbf{x})$, we see that $D_{j}f_{i}$ is the derivative of $f_{i}$ with respect to $x_{j}$, keeping the other variables fixed. The notation
$$
\frac{\partial f_{i}}{\partial x_{j}}
$$
is therefore used in place of $D_{j}f_{i}$, and $D_{j}f_{i}$ is called a partial derivative.
Theorem 3.7. existence of partial derivatives [br917]
Suppose $\mathbf{f}$ maps an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, and $\mathbf{f}$ is differentiable at a point $\mathbf{x} \in E$. Then the partial derivative $(D_{j}f_{i})(\mathbf{x})$ exist, and
$$
\mathbf{f}'(\mathbf{x})\mathbf{e}_{j} = \sum_{i=1}^{m}(D_{j}f_{i})(\mathbf{x})\mathbf{u}_{i} \qquad (1 \leq j \leq n).
$$
If $\mathbf{F}(\mathbf{x})$ is a scalar-valued function, then
$$D \mathbf{F}(\mathbf{x}) = \begin{bmatrix}D_{1}\mathbf{F} & D_{2}\mathbf{F} & \cdots & D_{n}\mathbf{F}\end{bmatrix}$$
It’s also called the gradient of $\mathbf{F}$.
Suppose $\mathbf{f}$ maps a convex open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$, $\mathbf{f}$ is differentiable in $E$, and there is a real number $M$ such that
$$
\lVert \mathbf{f}'(\mathbf{x}) \rVert \leq M
$$
for every $\mathbf{x} \in E$. Then
$$
\lvert \mathbf{f}(\mathbf{b}) - \mathbf{f}(\mathbf{a}) \rvert \leq M \lvert \mathbf{b} - \mathbf{a} \rvert
$$
for all $\mathbf{a} \in E, \mathbf{b} \in E$.
A differentiable mapping $\mathbf{f}$ of an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$ is said to be continuously differentiable in $E$ if $\mathbf{f}'$ is a continuous mapping of $E$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.
More explicitly, it is required that to every $\mathbf{x} \in E$ and to every $\epsilon > 0$ corresponds a $\delta > 0$ such that
$$
\lVert \mathbf{f}'(\mathbf{y}) - \mathbf{f}'(\mathbf{x}) \rVert < \epsilon
$$
if $\mathbf{y} \in E$ and $\lvert \mathbf{x} - \mathbf{y} \rvert < \delta$.
If this is so, we also say that $\mathbf{f}$ is a $\mathscr{C}'$-mapping, or that $\mathbf{f} \in \mathscr{C}'(E)$.
Theorem 3.10. necessary and sufficient conditions for $\mathbb{f} \in \mathscr{C}'(E)$ [br921]
Suppose $\mathbf{f}$ maps an open set $E \subset \mathbb{R}^{n}$ into $\mathbb{R}^{m}$. Then $\mathbf{f} \in \mathscr{C}'(E)$ if and only if the partial derivatives $D_{j}f_{i}$ exist and are continuous on $E$ for $1 \leq i \leq m, 1 \leq j \leq n$.
